The Laplace transform

Differential equations and Laplace transforms: Laplace-transformations

The Laplace transform

A known technique for solving (systems of) differential equations is based on the Laplace transformation. This is a linear mapping between certain vector spaces of functions. Here we discuss the definition of the Laplace transform and its linearity. Later we will discuss the most important calculation rules and applications to solving (systems of) differential equations.

Laplace transform

The Laplace transform of a function $f$ defined on $\ivco{0}{\infty}$ is the function $\laplace{(f)}$ , for sufficiently large $s$ defined by

$\laplace{(f)}(s) = \int_0^{\infty} f(t)\cdot \e^{-s\cdot t}\dd t$

The Laplace transformation is the mapping $\laplace{}$ which assigns to $f$ the Laplace transform $\laplace{(f)}$ of $f$ .

The Laplace transform of the constant function $1$ is

$\laplace{(1) }(s) =\int_0^{\infty} \e^{-s\cdot t}\dd t=\left[ \frac{-1}{s}\e^{-s\cdot t}\right]_{0}^{\infty}=0-\frac{-1}{s}=\frac{1}{s}$

that is defined on $\ivoo{0}{\infty}$ and so on each interval of form $\ivco{a}{\infty}$ where $a\gt0$ .

The Laplace transform thus assigns to a function $f$ on the interval $\ivco{0}{\infty}$ (for which the indicated definite integral exists) a function $\laplace{(f)}$ on an interval $\ivco{a}{\infty}$ . It is thus a mapping from the vector space of real-valued functions on $\ivco{0}{\infty}$ (for which the indicated definite integral exists) to the set $F$ of all the functions which are defined on an interval of the shape $\ivco{a}{\infty}$ .

Convention

The argument of the function $f$ is often denoted by $t$ in order to emphasize its association with time; this is related to the restriction $t\ge0$ . In the function rule of the Laplace-transform the variable $s$ is typically used as an argument. The Laplace transform then supplies a transition from the time domain ( $t$ -domain for short) to the frequency domain ( $s$ -domain for short).

Often, instead of $\laplace{(f)}(s)$ the notation $\laplace \{f(t) \}(s)$ or $\laplace \{f\}(s)$ is used. We avoid using the braces $\{$ and $\}$ , since their meaning is the usual one for brackets. If $f(t)$ is a compound expression, we also write $\laplace{(f(t))}(s)$ in order to make clear what the argument of $\laplace{}$ is. Note that in $\laplace{(f)}(s)$ the Laplace transform has priority over application to $s$ : it would be clearer if we wrote $\left(\laplace{(f)}\right)(s)$ , but we try and reduce the use of brackets so as to keep the expressions legible.

Sometimes, we use the notation $F(s)$ for $\laplace{(f)}(s)$ .

The Laplace transform is not defined for every function. But if $\laplace{(f)}(s_0)$ exists for a certain value $s_0$ , then it exists for all $s\ge s_0$ . After all, the function $\e^{-s\cdot t}$ of $t$ rapidly approaches $0$ if $t$ goes to $\infty$ ; the larger $s$ , the faster the approximation.

The Laplace transform is hardly ever calculated by use of the integral definition. Instead, calculation rules are used. The first such rule is linearity:

Laplace Linearity Theorem

Suppose that $f$ and $g$ are continuous functions with the property that $\laplace{(f)}(s)$ and $\mathcal{L}(g)(s)$ exist. If $\alpha$ , $\beta$ are real numbers, then $\laplace{(\alpha\cdot f+\beta\cdot g)} =\alpha\cdot\laplace{(f)}+\beta\cdot\laplace{(g)}$

In other words, the Laplacian operator is a linear map from the vector space of all continuous real functions for which the integral $\int_0^{\infty} f(t)\cdot \e^{-s\cdot t}\dd t$ exists, to the vector space of all real functions which are defined on an interval of the shape $\ivco{a}{\infty}$ for a real number $a$ .

$\begin{array}{rcl}\laplace{(\alpha\cdot f+\beta\cdot g)}(s) &=& \int_0^{\infty} (\alpha\cdot f+\beta\cdot g)\cdot \e^{-s\cdot t}\dd t\\&&\phantom{xx}\color{blue}{\text{definition Laplace transform}}\\&=& \int_0^{\infty} (\alpha\cdot f\cdot \e^{-s\cdot t}+\beta\cdot g\cdot \e^{-s\cdot t})\dd t\\&&\phantom{xx}\color{blue}{\text{distributivity of multiplication}}\\&=& \alpha \int_0^{\infty} f\cdot \e^{-s\cdot t}\dd t+\beta\cdot\int_0^{\infty}g\cdot \e^{-s\cdot t}\dd t\\&&\phantom{xx}\color{blue}{\text{linearity of integration}}\\ &=&\alpha\cdot\mathcal{L} f(s)+\beta\cdot\laplace{g}(s)\\&&\phantom{xx}\color{blue}{\text{definition of Laplace transform}} \end{array}$

Further calculation rules:

Calculation rules for Laplace transforms

Let $a$ and $b$ be real numbers with $b\gt0$ and let $n$ be a natural number.

$\begin{array}{lcrcl}\text{time scaling:}&\phantom{xx}&\laplace{\left(f(bt)\right)}(s) &=&\displaystyle\dfrac{1}{b} \laplace{f\left(\dfrac{s}{b}\right)}\\ \text{frequency shift:}&\phantom{xx}&\laplace{\left(\ee^{at}\cdot f(t)\right)}(s) &=& \laplace{(f)}(s-a)\\\text{derivative in frequency domain:}&\phantom{xx}&\laplace{\left(t^n\cdot f(t)\right)}(s) &=& (-1)^n\left(\laplace{(f)}\right)^{(n)}(s) \end{array}$

Special cases:

$\begin{array}{rcl} \laplace{\left(t\cdot f(t)\right)}(s) &=& -\left(\laplace{(f)}\right)'(s)\\ \laplace{\left(t^n\right)}(s) &=& \dfrac{n!}{s^{n+1}}\\ \laplace{\left(\ee^{at}\cdot \dfrac{t^{n-1}}{(n-1)!}\right)}(s) &=& \dfrac{1}{(s-a)^{n}}\end{array}$

As we will be able to see in the example below, the Laplace transforms of the trigonometic functions sine and cosine are given by

$\begin{array}{rcl}\laplace{\left(\cos(t)\right)}(s) &=&\displaystyle \frac{s}{s^2+1}\\ \laplace{\left(\sin(t)\right)}(s) &=&\displaystyle \frac{1}{s^2+1}\end{array}$

The rule for the time scaling gives

$\begin{array}{rclcl}\laplace{\left(t\cdot\cos(bt)\right)}(s) &=&\displaystyle \frac{1}{b}\left(\frac{\frac{s}{b}}{\left(\frac{s}{b}\right)^2+1}\right) &=& \displaystyle\frac{s}{s^2+b^2}\\ \laplace{\left(t\cdot\sin(bt)\right)}(s) &=&\displaystyle \frac{1}{b}\left(\frac{1}{\left(\frac{s}{b}\right)^2+1}\right)&=&\displaystyle\frac{b}{s^2+b^2}\end{array}$

The rule for the derivative in the frequency domain gives

$\begin{array}{rclcl}\laplace{\left(t\cdot\cos(t)\right)}(s) &=&\displaystyle- \frac{\dd}{\dd s}\left(\frac{s}{s^2+1}\right) &=& \displaystyle\frac{1-s^2}{(s^2+1)^2}\\ \laplace{\left(t\cdot\sin(t)\right)}(s) &=&\displaystyle -\frac{\dd}{\dd s}\left(\frac{1}{s^2+1}\right)&=&\displaystyle\frac{-2s}{(s^2+1)^2}\end{array}$

The rule for a frequency shift gives

$\begin{array}{rclcl}\laplace{\left(\e^{at}\cdot\cos(t)\right)}(s) &=& \displaystyle\laplace{\left(\cos(t)\right)}(s-a) &=& \displaystyle\frac{s-a}{(s-a)^2+1}\\ \laplace{\left(\e^{at}\cdot\sin(t)\right)}(s) &=&\displaystyle \laplace{\left(\sin(t)\right)}(s-a)&=&\displaystyle\frac{1}{(s-a)^2+1}\end{array}$

Time scaling:

$\begin{array}{rcl}\laplace{\left(f(bt)\right)}(s) &=&\int_0^{\infty}f(bt)\cdot \ee^{-st}\,\dd t\\&=&\displaystyle\frac{1}{b}\cdot\int_0^{\infty} f(u)\cdot \ee^{-\frac{s}{b}\cdot u}\,\dd u\\ &&\phantom{xx}\color{blue}{\text{substitution } t = \frac{u}{b}}\\ &=&\displaystyle \frac{1}{b}\laplace{(f)}\left(\frac{s}{b}\right)\\ \end{array}$

Frequency shift: $\begin{array}{rcl}\laplace{\left(\ee^{at}\cdot f(t)\right)}(s) &=&\int_0^{\infty}\ee^{at}\cdot f(t)\cdot \ee^{-st}\,\dd t\\&=&\int_0^{\infty} f(t)\cdot \ee^{-(s-a)\cdot t}\,\dd t\\&=&\laplace{(f)}(s-a)\\ \end{array}$

Derivative in frequency domain: we first prove the case $n=1$ ; here we use the fact that under the given circumstances, differentiation with respect to $s$ and integration with respect to $t$ may be interchanged changed

$\begin{array}{rcl} \laplace{\left(t \cdot f(t)\right)}(s) &=& \int_0^{\infty}t\cdot f(t)\cdot \ee^{-st}\,\dd t\\ &=&- \int_0^{\infty} \dfrac{\partial}{\partial s}\left(f(t)\cdot \ee^{-st}\right)\,\dd t\\ &=&- \dfrac{\dd}{\dd s} \int_0^{\infty} f(t)\cdot \ee^{-st}\,\dd t\\ &=&- \dfrac{\dd}{\dd s}\left(\laplace{(f)}(s)\right)\\ &=& -\left(\laplace{(f)}\right)'(s)\\ \end{array}$

By induction on $n$ we derive the general case, $n\gt 1$ , from this:

$\begin{array}{rcl} \laplace{\left(t^n\cdot f(t)\right)}(s) &=& \laplace{\left(t\cdot \left(t^{n-1}\cdot f(t)\right)\right)}(s)\\ &=&-\dfrac{\dd}{\dd s}\left(\laplace{(t^{n-1}f(t))}(s)\right)\\ &=&-(-1)^{n-1}\dfrac{\dd}{\dd s}\dfrac{\dd^{n-1}}{\dd s^{n-1}}\left(\laplace{(f(t))}(s)\right)\\ &=&(-1)^{n}\dfrac{\dd^{n}}{\dd s^{n}}\left(\laplace{(f(t))}(s)\right)\\ &=& (-1)^n\left(\laplace{(f)}\right)^{(n)}(s)\\ \end{array}$

The first two special cases follow from the rule for the derivative in the frequency domain by taking $n=1$ and $f(t)=1$ , respectively. The third special case follows from the previous special case by application of linearity and a frequency shift.

Here are some examples:

Compute the Laplace transform of the function $f$ on $\ivco{0}{\infty}$ defined by
$f(t) = 7 t+5$
Give your answer as a function in the variable $s$ for $s\gt 0$ .

$\laplace{(f)}(s) =$ ${{5 s+7}\over{s^2}}$

By use of the definition of Laplace transform we find
$\begin{array}{rcl}\laplace{(f)}(s) &=&\displaystyle \int_0^{\infty}\left(7 t+5\right)\cdot \ee^{-st} \,\dd t\\ &&\phantom{xx}\color{blue}{\text{definition of Laplace transform}}\\ &=&\displaystyle\lim_{T\to\infty} \int_0^T\left(7 t+5 \right)\cdot \ee^{-st} \,\dd t\\ &&\phantom{xx}\color{blue}{\text{definition of definite integral with upper bound }\infty}\\ &=&\displaystyle\lim_{T\to\infty}\left[-{{\left(7 s t+5 s+7\right) \euler^ {- s t }}\over{s^2}} \right]_{t=0}^{t=T}\\ &&\phantom{xx}\color{blue}{\text{antiderivative calculated}}\\ &=&\displaystyle\lim_{T\to\infty}\left({{5 s+7}\over{s^2}}-{{\left(7 T s+5 s+7\right) \euler^ {- T s }}\over{s^2}} \right)\\ &&\phantom{xx}\color{blue}{\text{antiderivative evaluated at boundary points}}\\ &=&\displaystyle {{5 s+7}\over{s^2}}-0\\ &&\phantom{xx}\color{blue}{\text{limit for }T\to\infty\text{ determined}}\\ &=&\displaystyle {{5 s+7}\over{s^2}} \end{array}$
By use of linearity we can find the Laplace transform more rapidly:
$\begin{array}{rcl}\laplace{(f)}(s) &=&\displaystyle \laplace{\left(7 t+5\right)}(s)\\ &&\phantom{xx}\color{blue}{\text{definition of }f}\\ &=&\displaystyle 7\cdot \laplace{\left( t\right)}(s)+5\cdot\laplace{\left(1\right)}(s)\\ &&\phantom{xx}\color{blue}{\text{linearity of the Laplace transform}}\\ &=&\displaystyle 7\cdot \frac{1}{s^2}+5\cdot\frac{1}{s}\\ &&\phantom{xx}\color{blue}{\text{Laplace transforms known from rules}}\\ &=&\displaystyle {{5 s+7}\over{s^2}} \end{array}$

New example